CHEMISTRY MOLE CONCEPT DALTON S ATOMIC THEORY By observing the laws of chemical combination, John Dalton proposed an atomic theory of matter. The main points of Dalton s atomic theory are as follows: (i) Matter is made up of extremely small, indivisible particles called atoms. (ii) Atom is the smallest particle that takes part in chemical reactions. (iii) Atoms of same substance are identical in all respect i.e. they possess same size, shape, mass, chemical properties etc. (iv) Atoms of different substances are different in all respect. (v) Atoms of different elements may combine with each other in a fixed, simple, whole number ratio to form compound atoms. (vi) Atom can neither be created nor destroyed. Limitation of Dalton s Atomic Theory (i) It failed to explain how atoms of different elements differ from each other. (ii) It failed to explain how and why atoms of elements combine with each other to form compound or molecules. (iii) It failed to explain the nature of forces that bind together different atoms in a molecule. (iv) It did not make any distinction between ultimate particle of an element that takes part in reaction (atoms) and the ultimate particle that has independent existence (molecules). MOLE The name Mole comes from Latin word for massive heap Mole is the unit, chemists use to keep track of large number of atoms, ions and molecules. One mole of object will always mean 6.023 10 23 of those objects. The number of object per mole, 6.023 10 23 mol 1 is called Avogadro s Number/Constant, represented by N A. Experimentally we know that mass of one carbon12 atom is 1.9926 10 23 gm. It follows that the number of atoms in exactly 12 gm of carbon 12 is 12 = 6.023 10 23 23 1.9926 10 This relationship of number of object to the atom is also true for ions or molecules. amu One atomic mass unit is defined as exactly 1/12 the mass of an atom of carbon12. We know experimentally that the mass of an atom of carbon12 is 1.9926 10 23 g, it follows that 1amu = 1.6605 10 24 g. The mass of any other type of atom in grams can be expressed in atomic mass unit by using this relation as a conversion factor.
PDT Courseware (FTRE)-10 th moving to 11 th -CH-2 GRAM ATOMIC MASS / GRAM MOLECULAR MASS The mass of one mole atoms of any element is exactly equal to the atomic mass in grams of that element. This is called as gram atomic mass or gram atoms. For example the atomic mass of oxygen atom is 16 amu. One mole of oxygen atoms contains 6.023 10 23 oxygen atoms. Now, mass of one atom of oxygen = 16 1.66 10 24 g Mass of one mole oxygen atom = 16 1.66 10 24 6.023 10 23 = 16 grams Similarly, the mass of 6.023 10 23 molecules (1 mole) of a substance is equal to its molecular mass in grams or gram molecule. So, mass of one mole of oxygen molecule O 2 = 32 1.66 10 24 6.023 10 23 or, 32 grams. So, 32 grams is one gram molecular mass or one gram molecule. As in atoms and molecules, mole concept is also applicable to ionic compound, which do not contain molecules. In such cases, the formula of any ionic compound is representation of ratio between constituent ions. One mole of an ionic compound is represented by 6.023 10 23 formula units. One mole of NaCl = 6.023 10 23 NaCl units = 6.023 10 23 units of Na + + 6.023 10 23 units of Cl Mass of one mole of NaCl = 23.0 g + 35.5 g = 58.5 gram NaCl MOLE AND MOLAR VOLUME From Avogadro s law it is known that equal volumes of all gases contain equal number of molecules under similar temperature and pressure conditions. Now, since one mole molecules of all gases contain same number (6.023 10 23 ) of molecules therefore, they occupy same volume under similar conditions of temperature and pressure. The volume occupied by one mole molecules of a gaseous substance is called Molar volume. One mole molecules of all gases occupy 22.4 litres at 273 K and 760 mm pressure (S.T.P.). Hence, molar volume of all gases at S.T.P. is 22.4 litres. MOLE AT A GLANCE For counting of articles, units like dozen or score is commonly used. Similarly in chemistry mole is the counting unit for chemical entities. 23 The amount of substance containing Avogadro s number 6.023 10 atom, molecule, ions, electrons and protons, i.e. (i) Number of moles of molecule Weight in gram = Molecular weight (ii) Number of moles of atoms Weight in gram = Atomic weight (iii) Number of moles of gases Volume at NTP = Standard molar volume (22.4 L) (iv) Number of milli moles Mole 1000 (v) For a compound P x Q y x moles of Q = y moles of P (vi) Number of entities Avogadro's number 6.023 10 23 Number of moles
PDT Courseware (FTRE)-10 th moving to 11 th -CH- 3 PERCENTAGE COMPOSITION AND MOLECULAR FORMULA Determination of chemical formula can be achieved by analysis the compound for the amounts of the elements (moles) in a given mass of the compound. eg. According to law of definite proportions 17.0 g of NH 3 always contains 14.0 g of N and 3.0 g of H. Mass % of N in NH 3 Mass % of H in NH 3 Mass of N in 1 mole of NH3 100 Mass of 1 mole of NH 3 100 = 17.65% 17 3 14 100 = 82.35% 17 EMPIRICAL AND MOLECULAR FORMULA If we know the percentage composition of the elements by chemical analysing we can calculate the relative number of atoms of each element in the molecule of the compound. This gives us the empirical formula of the compound. Further if the molecular mass is known then the molecular formula can easily be determined. Thus, while the empirical formula shows the relative number of atoms in the simplest ratio, the molecular formula gives the actual number of atoms of each element in a molecule. eg. A sample of hydrogen peroxide consists of 94.11% O and 5.89% H by mass. Therefore, the number of moles of each element in the 100 g sample is 1 94.11 5.88 mole of O 16 1 5.89 5.89 mole of H 1 Now, with help of the above data we can find out the number of moles of one element relative to the other. 5.89 mole of H 1 mole of H Thus, we have, 5.88 mole of O 1 mole of O Showing that for hydrogen peroxide, the ratio is 1 mole of hydrogen to 1 mol of oxygen and the empirical formula for hydrogen peroxide is HO. The molecular mass of hydrogen peroxide is 34 gmol 1 whereas the empirical formula mass for hydrogen peroxide is 17 gmol 1, therefore molecular formula for hydrogen peroxide is 34 2 times, the empirical formula. 17 i.e., (HO) 2 = H 2 O 2 STOICHIOMETRY A basic question raised in the chemical laboratory is, How much product will be formed from specific amounts of reactants? Or in some cases we might ask the reverse question: How much starting material (reactant) must be used to obtain a specific amount of product? To interpret a reaction quantitatively, we need to apply our knowledge of molar masses and the mol concept. For example, CaCO CaO g CO g 3 2 we see that CaCO 3 on decomposition gives CaO and CO 2. The above equation is balanced. But if the equation is not balanced, then first balance the equation. For above example, we can say that, 1 mol of CaCO 3 decomposes to give 1 mole of CaO and 1 mol of CO 2. or we can also say that, 100 gms of CaCO 3 gives 56 gm of CaO and 44 gms of CO 2. So, it it is asked that how much CaO will be formed from 20 gms of CaCO 3. So, we can apply unitary method and find out. Here, 100 gms CaCO 3 gives 56 gm of CaO.
PDT Courseware (FTRE)-10 th moving to 11 th -CH-4 20 gms CaCO 3 gives 56 20 11.2 gm 100 So, summarising the following points: The mole method consists of the following steps (i) Write correct formula for all reactants and products, and balance the resulting equation. (ii) (iii) (iv) Convert the quantities of some or all given or known substances (usually reactants) into mole. Use the coefficients in the balanced equation to calculate the number of mole of the unknown quantities (usually products). Using the calculated number of mole and molar masses, convert the unknown quantities to whatever units are required. Mole of reactant Mole of product Mass of reactant Mole of reactant Mole of product Mass of reactant Mole of reactant Mole of product Mass of product LIMITING REAGENT A limiting reactant in a reaction is the species supplied in an amount smaller than that required by the stoichiometric relation between the reactants. A limiting reactant is like a part in an automobile factory, if there are 1000 headlights and 600 car bodies, then the maximum member of automobile will be limited by the number of headlights. Because each body requires two headlights and there are enough headlights are for only 500 cars. So the headlights play the role of limiting reagents. MOLE FRACTION The mole fraction of any component in the solution is equal to the number of moles of that component divided by the total number of moles of all the components. For a solution containing n 2 moles of the solute dissolved in n 1 mole of the solvent, Mole fraction of solute in the solution n2 x2 n n 1 2 Mole fraction of solvent in the solution n1 x1 n n 1 2 x 1 + x 2 = 1
PDT Courseware (FTRE)-10 th moving to 11 th -CH- 5 A S S I G N M E N T P R O B L E M S 1. What is the mass of one mole of electrons? Mass of one electron is 9.11 10 31 kg. 2. Simplest formula of the compound containing 60 % of element A (At. wt.10) and 40% of element B (At.wt.20) is 3. What volume of CCl 4 having density 1.5 g/cc contains 1 10 25 chlorine atoms. 4. Calculate the number of atoms of oxygen present in 300 g of CaCO 3. 5. Calculate the mass of one atom of calcium, if atomic weight is 40 g mol 1. 6. The number of atoms and molecules of nitrogen in 224 ml of nitrogen at STP are x and y respectively. Calculate the value of x and y. 7. By heating 10 g CaCO 3, 5.6 g CaO is formed. What is the weight of CO 2 obtained in this reaction? 8. Find out empirical formula of an alkane (hydrocarbon) containing 82.76% carbon by weight? 9. An inorganic salt gave the following percentage composition: Na = 29.11, S = 40.51 and O = 30.38. Calculate the empirical formula of the salt. 10. Find out the mole of each component in 48 g of Na 2 CO 3. 11. How many atoms are there in 1 gm of CaCO 3? 12. The molecular formula of ferric sulphate is Fe 2 (SO 4 ) 3. How many moles are there in 40 g of ferric sulphate? (Atomic mass: Fe = 56, S = 32, O = 16) 13. If 200 g of calcium carbonate (formula weight = 100) occupies a volume of 67.0 ml, what is its density? 14. Calculate the molecular mass of (NH 4 ) 2 SO 4.FeSO 4.6H 2 O. 15. Calculate the number of moles present in 5 g of Ca. 16. What will be the volume of 1 gram Helium at NTP? 17. Calculate the number of molecules of sulphur (S 8 ) in 16 g of solid sulphur. 18. How many grams of neon will have the same number of atoms as are there in 4 g of Ca? Atomic mass of Ne = 20; Ca = 40 amu] 19. Calculate the number of moles of 52 g of He. 20. Calculate the number of moles of 12.044 10 23 number of He atoms. 21. Which of the following samples contains 2.0 10 23 atoms? (a) 8.0 g O 2 (b) 3.0 g Be (c) 8.0 g C (d) 19.0 g F 2
PDT Courseware (FTRE)-10 th moving to 11 th -CH-6 22. Which of the following samples contains the largest number of atoms? (A) 1 g of Ni(s) (B) 1 g of Ca(s) (C) 1 g of N 2 (g) (D) 1 g of B(s) 23. Mass of single atom of X is 2.65 10 23 g. X should be (A) 12 C (B) 14 N (C) 16 O (D) 1 H 24. Moles of He constitute by 9.033 10 24 atoms of He is 25. The volume of 2 moles of CO 2 gas at NTP is equal to.litres. 26. Which of the following has maximum number of atoms? (A) 24 g of C(12) (B) 56 g of Fe(56) (C) 27 g of Al(27) (D) 108 g of Ag(108) 27. If a mole were to contain 1 10 24 particles, the mass of one mole of oxygen is (in g) 28. 1.12 ml of a gas is produced at STP by the action of 4.12 mg of alcohol, ROH with methyl magnesium iodide. The molecular mass of alcohol is ROH CH3MgI CH4 R OMgI 29. 2 mol of H 2 S and 11.2 L SO 2 at N.T.P. reacts to form x mol of sulphur; x is SO 2 + 2H 2 S 3S + 2H 2 O 30. Hydrogen evolved at NTP on complete reaction of 27 gm of Al with excess of aq. NaOH would be (Chemical reaction: 2Al + 2NaOH + 2H 2 O 2NaAlO 2 + 3H 2 )
PDT Courseware (FTRE)-10 th moving to 11 th -CH- 7 A N S W E R S T O A S S I G N M E N T 1. 54.8422 10 8 kg 2. A 3 B 2 3. 0.426 litre 4. 5.42 10 24 5. 6.64 10 23 g 6. 1.210 22 atoms, 6.023 10 21 molecules 7. 4.4 gm 8. C 2 H 5 9. Na 2 S 2 O 3 10. Na = 1, C = 0.5, O = 1.5 11. 3.012 10 22 12. 0.1 13. 2.99 g/ml 14. 392 15. 0.125 16. 5.6 litre 18. 2g 19. 13 20. 2 21. b 22. D 23. C 24. 15 25. 44.8 26. A 27. 53.2 28. 82.4 29. 1.5 30. 33.6 litre